Integration sikhao!!!

Strider koi question pooch
usse samajhenge[1]

6)

(5-4x-x²)^5/2 = (5+x)^5/2(1-x)^5/2

now write numerator as : 1/6[(5+x) + (1-x)]

then the integration bcmes:

1/6 [ ∫(5+x)/(5+x)^5/2(1-x)^5/2 dx + ∫(1-x)/(5+x)^5/2(1-x)^5/2 dx. ]

=1/6[ ∫dx/(5+x)^3/2(1-x)^5/2] + ∫dx/(5+x)^5/2(1-x)^3/2 ]

now apply the trick of maverick...

remeber that is a standard trick ...

PLEASE COMPARE THE QUESTIONS WITH JEE LEVEL OF TOUGHNESS AND THEN SOLVE:

#5)

(Subjective again)

\int \frac{(x-1)\; dx}{(x+1)\sqrt{x^3+x^2+x}}

#6)

\int \frac{dx}{\left(5-4x-x^2 \right)^{\frac{5}{2}}}

#7)

\int \frac{dx}{(x-1)^\frac{3}{4}(x+2)^\frac{5}{4}}}

#8)

Prove that

\int_{0}^{1}{\frac{tan^{-1}x\; dx}{x}}=\int_{0}^{\frac{\pi }{2}}{\frac{y\; dy}{siny}}

[1]

Thanx da

So everytime, i get a question like this...... i'll remember u da. [1]

YUP ARAGON I THINK U GOT IT NOW

FOR UR INFORMATION U SOLVED A QUES OF IIT 1997 [1]

that'll become......

\left[\sqrt{1+t^2} \right]^{2}_{0}

Hence....(√5-1)??

SOLVE THE FULL QUESTION
UWIONT GET ANYTHING FOR HALF SOLUTION[2]

DO IT THEN WE WILL CHECK[1]

putting r/n---->t, dr/n----dt

It became
\int_{0}^{2}{\frac{t\; dt}{\sqrt{1+t^2}}}

Correct till now??

\lim_{n->infinity} 1/n\sum_{r=1}^{2n}{}r/\sqrt{n^{2}+r^{2}}

THIS ONE IS FOR ANIRUDH ONLY!!!!!!!!

yes i may say
in most of the other cases u have to struggle hard to find it
but it of premeir importance
chal i give u a ques 4 practice

What should we look to do in such questions??

Always substitute r/n--->t??

yaar dekh iss mein
put r/n------>t
then 1/n-------->dt

this could be represented in the form of integral
lower limit and upper limit can be found out in the following method

For lower limit
sigma ke nicche waali term check kar
aggar woh constant hai to u put lower limit as 0

For upper limit
check the coefficient of n in the upper power of sigma
the coefficient must be written in the upper limit

so

noe the ques becomes

∫₀¹x⁹⁹dx

and hence 1/100
[1]

Pls post complete solution with easy-to-understand reasoning

Exclude terminologies as much as possible

OK then....

\lim_{n\rightarrow \infty}\frac{1}{n^{100}}\sum_{r=1}^{n}{r^{99}}

(II)

Subjective type

(1)\int sin^3x\; cos(\frac{x}{2})dx

(2)\int sin^{-1}\left(\sqrt{\frac{x}{a+x}} \right)dx

(3)\int \frac{dx}{sinx(3+2cosx)}

(4)\int \frac{sinx\; dx}{sin3x}

Pls give me the algorithm for converting summation to a definite integral...

Also pls illustrate with an example!

[11] [11] [11] [11] [11] [11] [11]

@manipal.... afetr where i left it .. in 2nd questions...

its jus one step...

becoz tanθsec²Î¸dθ taken as second function... a child can integrate..

anyways... u have given another method... dat way its nice :)

subj. 3)

∫dx/sinx(3+2cosx)

= ∫sinxdx/sin²x(3+2cosx)

= ∫sinxdx/(1-cos²x)(3+2cosx)

now put cosx=t...

partial fractions??

GUESSED SO!![3] [3]

Q2
I THINK IT WOULD BE BETTER TO PUT
x+a=t²
dx=2tdt

upper niche waala t kat jaaega
so the ques becomes
∫sin^-1√t²-adt
this could be found out by PARTS (EDITED)[1]

4) ∫sinx/sin3x = ∫sinx/sinx[3-sin2x] = ∫1/[2+cos²x]

=∫sec²xdx/[2sec²x+1]

=∫sec²xdx/[3+2tan²x]

tanx =t le lena ......

then it becomes simple quadratic..

subjective:

2) sin^-1√x/(x+a)

take x=atan²Î¸ => dx = 2atanθsec²Î¸dθ

∫sin^-1√x/(x+a)

=∫sin^-1(sinθ) .2atanθsec²Î¸dθ

= 2a∫θtanθsec²Î¸dθ

now do partial fractions taking θ as first function and tanθsec²Î¸ as second function.

final ans : a[((x/a) +1)tan^-1√x/a - √x/a ]

for 1 subjective problem put sinx=2sinx/2 cosx/2 n then put cos x/2 as t